Question

我想要3张桌子：

书籍（id，title，auto increment（id））
作者（id，name，auto increment（id））
book_author（book_id，author_id）

我想在一个事务中创建所有内容，以确保只有在所有表中一切正常时才会将数据插入到表中。

要向book_author插入任何内容，我需要id，而我没有。

我应该：

将数据插入books表，进行选择以获取ID，
将数据插入author表，进行选择以获取ID，
将两个id插入book_author

或者我应该在插入时使用触发器？但这是一种使触发器依赖于两个插入的方法吗？

我会帮忙，我可以说我正在尝试使用python和sqlalchemy。

第二个问题也许这很重要......如果某个数据库不支持自动增量怎么办？

Answer 1

SQLAlchemy Object Relation API为您完成所有工作：添加记录，支持事务，自动增加列值。因此，您不需要为每个对象保存id并为每个关系创建记录。用例子展示这个的最好方法是：

from sqlalchemy import *
from sqlalchemy import create_engine, orm
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship

__author__ = 'vvlad'

metadata = MetaData()
Base = declarative_base()
Base.metadata = metadata


"""many-to-many relationship table"""
author_book = Table(
    'author_book', Base.metadata,
    Column('author_id',String(11),ForeignKey('author.id')),
    Column('book_id',Integer,ForeignKey('book.id')),
)

class Author(Base):
    __tablename__ = 'author'
    """SQLAlchemy's Sequence(name) for Column allows to have database agnostic autoincrement"""
    id = Column(Integer, Sequence('author_seq'), primary_key=True)
    name = Column(String(length=255))
    def __repr__(self):
        return "%s(name=\"%s\",id=\"%s\")" % (self.__class__.__name__,self.name,self.id)


class Book(Base):
    __tablename__ = 'book'

    id = Column(Integer, Sequence('book_seq'),primary_key=True)
    name = Column(String(length=255))
    """Defines many-to-many relations"""
    authors = relationship(Author,secondary=author_book,backref="books",collection_class=list)


    def __repr__(self):
        return "%s(name=\"%s\",id=\"%s\")" % (self.__class__.__name__,self.name,self.id)




db = create_engine('sqlite:////temp/test_books.db',echo=True)

#making sure we are working with a fresh database
metadata.drop_all(db)
metadata.create_all(db)

sm = orm.sessionmaker(bind=db, autoflush=True, autocommit=True, expire_on_commit=True)
session = orm.scoped_session(sm)

"""associating authors with book"""
b1 = Book(name="Essential SQLAlchemy")
a1 = Author(name="Rick Copeland")
"""associating book with author"""
a1.books.append(b1)
b2 = Book(name="The Difference Engine")
gibson = Author(name="William Gibson")
b2.authors.append(gibson)
b2.authors.append(Author(name="Bruce Sterling"))

b3 = Book(name="Pattern Recognition")
b3.authors.append(gibson)
try:
    #here your transaction start
    #adding objects to session. SQLAlchemy will add all related objects into session too
    session.add(b2)
    session.add(a1)

    session.add(b3)
    #closing transaction
    session.commit()
#remember to put specific exceptions instead of broad exception clause
except:
    """if something went wrong - rolls back session (transaction)"""
    session.rollback()

print "Book info"
b3 = session.query(Book).filter(Book.name=="Essential SQLAlchemy").one()
print b3
for author in b3.authors:
    print author


aname = "William Gibson"
print "Books of author %s" % aname
for book in session.query(Book).join(author_book).join(Author).filter(Author.name==aname).all():
    print book
    for author in book.authors:
        print author

Answer 2

如果您将大量记录保存到表中，则在使用SELECT语句创建记录后查找记录的ID可能会有问题。使用触发器可以轻松引用插入的最后一条记录。这是一个tutorial。

触发器还可以帮助您自动递增表格的ID字段。您可以看到示例here。

经典问题：如何将数据插入表格：书籍，作者，书籍作者？

2 个答案: